ABCDEF - All Combinations of 3 letter word I'm not sure to how to correctly solve it, because I've never learned combinatorics.
At first I thought how many combinations of 3 letter word are possible, so I think it is $\binom{6}{3}=120$, then in each word I get I can change the order of the letters so it will be $3!$. So is the answer $\binom{6}{3}\cdot3!=720$? Is the way ok? 
Then if I want to add the words that can have the same letters 2-3 times, what then I do?
 A: The answer that you have given is ok except $\binom{6}{3}\cdot3!=120$ (because $\binom{6}{3} = 20$). Your other question's answer is $6^3 = 216$ because for each of the three letter, you have $6$ choice.
A: If letters cannot repeat in a three-letter word, there are $6$ choices (A, B, C, D, E, or F) for the first letter. There are then $5$ choices for the second letter (the five letters that were not chosen in the first letter) and then there are $4$ choices for the third letter. This gives $6 \cdot 5 \cdot 4 = 120$ three-letter words with repeats not allowed.
If letters can be repeated as many times as you want, there are $6$ options (A, B, C, D, E, or F) for the first letter, second letter, and third letter. Then $6^3 = 216$ are the number of options for all three-letter-words.
If a letter can be repeated at most twice, it gets more complicated. Notice that a three-letter word has all different letters, two letters that are the same and one that is different, or all three letters the same. Without any restrictions on the number of repetitions, we found $216$ three-letter words. Of those, there are exactly $6$ which have letters repeated $3$ times (AAA, BBB, CCC, DDD, EEE, and FFF). That means the other $216 - 6 = 210$ three-letter-words have letters repeated at most twice.
A: Yes for the first part it is correct $$\binom{6}{3}\cdot3!=120$$
For the second part if you want consider for example also the case with 2 A you have to compute
$$\binom{6}{3}\cdot3!+5\cdot 3=135$$
and if you want consider also the case with 3 A you have to compute
$$\binom{6}{3}\cdot3!+5\cdot 3+1=136$$
Thus if you want consider for example also the case with 2 letters
$$\binom{6}{3}\cdot3!+6\cdot5\cdot 3=210$$
and if you want consider also the case with 3 letters you have to compute
$$\binom{6}{3}\cdot3!+6\cdot5\cdot 3+6=216=6^3$$
A: If you want $3$ letter words over the alphabet $A, B, C, D, E, F$, where each letter can appear as much as want in the word, then there are
$$
6\times6\times 6
$$
such words. 
